How to zip two differently sized lists?

I want to zip two list with different length

for example

A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]

and I expect this

[(1, 'A'), (2, 'B'), (3, 'C'), (4, 'A'), (5, 'B'), (6, 'C'), (7, 'A'), (8, 'B'), (9, 'C')]

But the build-in zip won't repeat to pair with the list with larger size . Does there exist any build-in way can achieve this? thanks

here is my code

idx = 0
zip_list = []
for value in larger:
    zip_list.append((value,smaller[idx]))
    idx += 1
    if idx == len(smaller):
        idx = 0

You can use itertools.cycle:

Make an iterator returning elements from the iterable and saving a copy of each. When the iterable is exhausted, return elements from the saved copy. Repeats indefinitely.

Example:

A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]

from itertools import cycle
zip_list = zip(A, cycle(B)) if len(A) > len(B) else zip(cycle(A), B)

Try this.

A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
Z = []
for i, a in enumerate(A):
    Z.append((a, B[i % len(B)]))

Just make sure that the larger list is in A.


Do you know that the second list is shorter?

import itertools
list(zip(my_list, itertools.cycle(another_list)))

This will actually give you a list of tuples rather than a list of lists. I hope that's okay.


Solution for an arbitrary number of iterables, and you don't know which one is longest (also allowing a default for any empty iterables):

from itertools import cycle, zip_longest

def zip_cycle(*iterables, empty_default=None):
    cycles = [cycle(i) for i in iterables]
    for _ in zip_longest(*iterables):
        yield tuple(next(i, empty_default) for i in cycles)

for i in zip_cycle(range(2), range(5), ['a', 'b', 'c'], []):
    print(i)

Outputs:

(0, 0, 'a', None)
(1, 1, 'b', None)
(0, 2, 'c', None)
(1, 3, 'a', None)
(0, 4, 'b', None)

You can use itertools.cycle:

from itertools import cycle

my_list = [1, 2, 3, 5, 5, 9]
another_list = ['Yes', 'No']

cyc = cycle(another_list)

print([[i, next(cyc)] for i in my_list])
# [[1, 'Yes'], [2, 'No'], [3, 'Yes'], [5, 'No'], [5, 'Yes'], [9, 'No']]